Consider a $m\times n$ matrix $A$ prove that: \begin{equation} \max_{x\in S^{n-1}}\|Ax\|_2=\max_{x\in S^{n-1}, y\in S^{m-1}}\langle Ax,y\rangle \end{equation}
MY ATTEMPT
$(\leq)$ Recall that if we set $y=\frac{Ax}{\|Ax\|_2}$ we have that $y\in S^{m-1}$; then: \begin{equation} \max_{x\in S^{n-1}, y\in S^{m-1}}\langle Ax,y\rangle\geq\max_{x\in S^{n-1}}\langle Ax,\frac{Ax}{\|Ax\|_2}\rangle =\max_{x\in S^n-1}\frac{\|Ax\|_2^2}{\|Ax\|_2}=\max_{x\in S^n-1}\|Ax\|_2 \end{equation}
Can someone give me an hint to verify the opposite inequalitY?
$S^{n-1}$ being compact and $x\rightarrow\|Ax\|_{2}$ is a continuous map, so it attains the max, say, \begin{align*} \max_{x\in S^{n-1}}\|Ax\|_{2}=\|Ax_{0}\|_{2} \end{align*} for some $x_{0}\in S^{n-1}$. If $\|Ax_{0}\|_{2}=0$ the result holds immediately. Now we set $y_{0}=Ax_{0}/\|Ax_{0}\|_{2}$, then $y_{0}\in S^{n-1}$ and hence \begin{align*} \|Ax_{0}\|_{2}=\sqrt{\left<Ax_{0},Ax_{0}\right>}=\sqrt{\left<Ax_{0},y_{0}\right>}\sqrt{\|Ax_{0}\|_{2}}, \end{align*} and hence \begin{align*} \|Ax_{0}\|_{2}=\left<Ax_{0},y_{0}\right>\leq\max_{x,y\in S^{n-1}}\left<Ax,y\right>. \end{align*}
For the other side, we apply Cauchy-Schwarz inequality, so \begin{align*} \left<Ax,y\right>\leq\left|\left<Ax,y\right>\right|\leq\|Ax\|_{2}\|y\|_{2}, \end{align*} when $y\in S^{n-1}$, then $\|y\|_{2}=1$, the rest is clear then.